A thin uniform rod of length 'L' and mass 'M' is bent at the middle point ' 0 ' at an angle of $45^{\circ}$…

A thin uniform rod of length 'L' and mass 'M' is bent at the middle point ' 0 ' at an angle of $45^{\circ}$ as shown in the figure. The moment of inertia of the system about an axis passing through '0' and perpendicular to the plane of the bent rod, is
  1. $\frac{\mathrm{ML}^{2}}{12}$
  2. $\frac{M L^{2}}{24}$
  3. $\frac{\mathrm{M} \mathrm{L}^{2}}{3}$
  4. $\frac{\mathrm{M} L^{2}}{6}$

Solution

Moment of inertia of each half of the rod about the mid point is given by $I_{1}=\frac{\frac{M}{2}\left(\frac{L}{2}\right)^{2}}{3}=\frac{M L^{2}}{24}$ Total moment of inertia $=\mathrm{I}=2 \mathrm{I}_{1}=\frac{\mathrm{ML}^{2}}{12}$

Asked in: MHT CET 2020 (14 Oct Shift 2)

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